In(ten mm, S/1000),(23)Appl. Sci. 2021, 11,8 of2.three. Calculation in the New 1-?Furfurylpyrrole Description hanger Installation Course of action The installation with the new hanger is basically the reverse course of action of the hanger removal. Nonetheless, the tension course of action for the duration of the installation with the new hanger will be the same as that with the unloading approach, because the pocket hanging hanger is carried out by way of the jack pine oil without the need of the need to reduce it. 2.three.1. Initial State The initial state would be the state just before the new hanger is installed: (a) New hanger: elasticity modulus is En , cross-sectional location is an , and cable length s is L0 . (b) Pocket hanging hanger: elasticity modulus is E , cross-sectional area is actually a , cable s s s length is L 0 , shear force is Q0 , and cable tension is T0 . Because the new hanger is installed following the old hanger is removed, then there is: L0=Ls d N, s T0 = TN , g(24)In accordance with the displacement coordination and force balance, it has:s L0 = s T0 L 0 s + L 0, EA s(25) (26)s s T0 = Q0 + G,two.three.2. The ith(i = 1, two, . . . , Nn ) Times Tension of the New Hanger Immediately after the ith occasions tension in the new hanger, let the new hanger internal force be Fiz , the pocket hanging hanger internal force be Tiz , the unstressed lengths of your new hanger z z and pocket hanging hanger be Li , L i , respectively, as well as the displacement on the ith instances tension with the new hanger be xiz . There isn’t any distinction between this process along with the ith instances of the pocket hanging; for that reason, the derivation will not be repeated and you can find:s Tiz = Tis-1 – E A Tis-1 – G – Qi-1 + Fiz z , z Li(27) , (28) (29)=En An Ls i -s s s G + Qi-1 – Fiz – Tis-1 z + En An Li-1 + Li-1 Fis-Fiz + En An xiz = Ls i -1 s Tis-1 – G – Qi-1 + Fiz z , s i -1 k +where z = 1/ LEA .2.3.three. The ith(i = 1, 2, . . . , Nn ) Occasions Unloading on the Pocket Hanging Hanger Just after the ith occasions unloading of your pocket hanging hanger, let the new hanger internal force be Fis , the pocket hanging hanger internal force be Tis , the unstressed lengths on the s s new hanger and pocket hanging hanger be Li , L i , respectively, plus the displacement of your ith instances tension of the new hanger be xis .z Fis = Fiz – En An ( Fiz – G – Qi + Tis ) s ,(30)zLi =sz z E A Li G + Qi – Tis – Fiz s + E A L i + L i Tiz , Tis + E A z z xis = Li ( Fiz – G – Qi + Tis ) s ,z(31) (32)z where s = 1/ Li k + En An .= = – – + ,Appl. Sci. 2021, 11,,(31) (32)9 ofwhere = 1/ + .two.three.four. Displacement Control two.3.4. Through the above calculation, it may be observed that following the ith = 1, 2, … , times Displacement Handle By means of the above calculation, it may be noticed that immediately after the the = 1, end . , Nn occasions tension with the new hanger, the accumulative displacement ofith (ilower 2, . . on the)hanger tension on the new hanger, the accumulative displacement with the reduced finish on the hanger is: is:- = – 1 i1 z +s + , Xiz = (i – 1) n=1 ( xn + xn ) + xiz ,(33) (33)After the ith = 1, two, … , times unloading of the pocket hanging hanger, the acAfter the ith (i = 1, two, . . . , Nn ) occasions unloading of your pocket hanging hanger, the cumulative displacement with the reduced finish of the hanger to become replaced is: accumulative displacement Xis with the reduce end of your hanger to be replaced is: = + , (34) i s z s Xi = n = 1 ( x n + x n ) , (34) , , and manage displacement threshold [D] really need to satisfy the following relationship: iz , Xis , and handle displacement threshold [D] must satisfy the following connection: X [], g [], Xid [ D ], Xi [.
